1. ## help! trig identity

I usually don't have any problem with these, but this one has me stuck.

1 - sinA = (sin(A/2) - cos(A/2))^(2)
subbing in the half-angle identities
1 - sinA = (((1 - cosA) / 2)^(1/2) - ((1 + cosA) / 2)^(1/2))^(2)
adding the complex fractions to simplify
1 - sinA = (((1 - cosA)^(1/2) - (1 + cosA)^(1/2)) / 2)^(2)
squaring (FOIL)
1 - sinA = ((1 - cosA) - (1 - cosA) - (1 - cosA) + (1 - cosA))/2
simplify
1 - sinA = 2cosA/2
not true
1 - sinA = cosA

It is tempting to say cos^(2)A = 1 - sin^(2)A, so cosA = 1 - sinA, but last time I checked, it does not work like that. CosA = (1 - sin^(2)A)^(1/2), and since that is a false square -- (1 - sinA)(1 + sinA) -- there is no real way to take the root. Tell me where I am wrong.

2. Originally Posted by sinewave85
I usually don't have any problem with these, but this one has me stuck.

1 - sinA = (sin(A/2) - cos(A/2))^(2)

Expand the square:

1 - sin(A) = sin^2(A/2) -2 sin(A/2)cos(A/2) + cos^2(A/2)

............. = 1 - 2sin(A/2)cos(A/2),

so you need only show:

sin(A) = 2sin(A/2)cos(A/2),

which I hope looks familiar.

RonL

3. Hello, sinewave85!

The Captain's method is the way to go, but your approach will work, too.

$\displaystyle 1 - \sin A \:= \:\left[ \sin\left(\frac{A}{2}\right) - \cos\left(\frac{A}{2}\right) \right]^2$

Using half-angle identities, the right side becomes: .$\displaystyle \left[\sqrt{\frac{1-\cos A}{2}} - \sqrt{\frac{1+\cos A}{2}}\right]^2$

Square: . $\displaystyle \frac{1-\cos A}{2} \:- \:2\cdot\sqrt{\frac{1-\cos A}{2}}\cdot\sqrt{\frac{1-\cos A}{2}} \:+ \:\frac{1 + \cos A}{2}$

We have: .$\displaystyle \left(\frac{1-\cos A}{2} + \frac{1+\cos A}{2}\right) \:- \:2\!\cdot\!\sqrt{\frac{1-\cos^2A}{4}}$

. . . . . . $\displaystyle =\;1 \:- \:2\!\cdot\!\sqrt{\frac{\sin^2A}{4}}$

. . . . . . $\displaystyle = \;1 \:- \:2\!\cdot\!\frac{\sin A}{2}$

. . . . . . $\displaystyle \;=\;1 - \sin A$ . . . ta-DAA!

4. Thanks for the help!